# How show the number is negative integer?number` (cos(pi/4)+(-1)^.5sin(pi/4))^100` ?

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### 1 Answer

You need to use the complex number theory to substitute `i` for `(-1)^(0.5)` such that:

`(cos(pi/4) + i*sin(pi/4))^100`

Using De Moivre's identity yields:

`(cos(pi/4) + i*sin(pi/4))^100 = (cos((100pi)/4) + i*sin((100pi)/4))`

`(cos(pi/4) + i*sin(pi/4))^100 = cos 25pi + i*sin 25pi`

`cos 25 pi = cos(24pi + pi)`

Using the trigonometric identity `cos(a+b) = cos a*cos b - sin a*sin b` yields:

`cos(24pi + pi) = cos 24pi* cos pi - sin 24pi*sin pi`

Since `sin pi = 0` and `cos pi = -1` yields:

`cos(24pi + pi) = -cos 24pi `

Using the trigonometric identity `sin(a+b) = sin a*cos b - cos a*sin b` yields:

`sin(24pi + pi) = sin 24pi* cos pi + cos 24pi*sin pi`

`sin(24pi + pi) = -sin 24pi`

`cos 25pi + i*sin 25pi =-cos 24pi -i* sin 24pi`

`sin 24pi = sin 2*12 pi = 2 sin 12 pi*cos 12 pi`

`sin 12 pi = sin 2*6pi = 2 sin 6pi*cos 6pi`

`sin 6pi = 2 sin 3pi*cos 3pi`

`sin 3pi = sin (2pi + pi) = 0 => sin 6pi = 0 => sin 12 pi = 0 => sin 24pi = 0`

`cos 24pi = cos^2 12 pi - sin^2 12 pi`

` sin^2 12 pi = 0 => cos 24pi = cos^2 12 pi`

`cos 12 pi = cos^2 6pi`

`cos 6pi = cos^2 3pi`

`cos 3pi = cos (2pi + pi) = -1 => cos^2 3pi = (-1)^2 = 1 => cos 12 pi = 1 => cos 24pi = 1`

`-cos 24pi -i* sin 24pi = -1 + i*0 => -cos 24pi -i* sin 24pi = -1`

Since `(cos(pi/4) + i*sin(pi/4))^100 = -cos 24pi -i* sin 24pi => (cos(pi/4) + i*sin(pi/4))^100 = -1` .

**Hence, checking if the expression `(cos(pi/4) + i*sin(pi/4))^100` is an integer negative, using De Moivre's identity, yields `(cos(pi/4) + i*sin(pi/4))^100 = -1` (integer negative).**